AMC 10 · 2002 · #4
Grade 6 arithmeticWhat is the value of (3x−2)(4x+1)−(3x−2)4x+1 when x=4?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Find the value of $(3x-2)(4x+1) - (3x-2)\cdot4x + 1$ when $x=4$.
Givens: The expression $(3x-2)(4x+1) - (3x-2)\cdot4x + 1$; The first two terms both contain the piece $(3x-2)$; the final $+1$ stands alone outside both products; $x=4$; Answer choices: (A) $0$, (B) $1$, (C) $10$, (D) $11$, (E) $12$
Unknowns: The single number the expression equals when $x=4$
Understand
Restated: Find the value of $(3x-2)(4x+1) - (3x-2)\cdot4x + 1$ when $x=4$.
Givens: The expression $(3x-2)(4x+1) - (3x-2)\cdot4x + 1$; The first two terms both contain the piece $(3x-2)$; the final $+1$ stands alone outside both products; $x=4$; Answer choices: (A) $0$, (B) $1$, (C) $10$, (D) $11$, (E) $12$
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Both of the first two terms carry the same factor $(3x-2)$, so instead of multiplying everything out, Tool #15 (Organize Information in More Ways) rewrites the expression to pull that shared factor to the front: $(3x-2)\big[(4x+1)-4x\big]+1$. The bracket collapses to a single number, which turns a messy problem into a one-line one. Tool #7 (Identify Subproblems) then handles the leftover in tidy steps — simplify the bracket, add the stray $+1$, substitute $x=4$. Tool #3 (Eliminate Possibilities) guards the built-in traps: dropping the outside $+1$ gives $10$ (kills (C)), and mishandling the leftover constants $-2+1$ as $0$ gives $12$ (kills (E)).
Execute — Answer: D
6.EE.A.3 Step 1 Pull out the shared factor
- Look at the first two terms: $(3x-2)(4x+1)$ and $(3x-2)\cdot4x$.
- Both are $(3x-2)$ times something.
- By the distributive property run in reverse (factoring), $(3x-2)(4x+1)-(3x-2)\cdot4x = (3x-2)\big[(4x+1)-4x\big]$.
- The lone $+1$ at the very end is not part of either product, so carry it along unchanged.
💡 When two terms share a factor, you can lift it out front and subtract what is left inside.
6.EE.A.3 Step 2 Simplify the bracket
- Inside the bracket, subtract: $(4x+1)-4x$.
- The $4x$ and $-4x$ cancel, leaving just $1$.
- So the two big products together equal $(3x-2)\cdot1 = 3x-2$.
- All the $4x$ machinery has vanished.
💡 Adding then subtracting the same $4x$ leaves nothing behind, so the bracket is just $1$.
6.EE.A.3 Step 3 Add the leftover $+1$
- Now bring back the $+1$ that was waiting outside: $3x-2+1$.
- Combine the constants $-2+1=-1$.
- The whole expression simplifies to $3x-1$.
- Do not drop that outside $+1$ — leaving it off is the mistake that produces $3x-2$.
💡 The stray $+1$ nudges the constant from $-2$ up to $-1$; it must not be forgotten.
6.EE.A.2 Step 4 Substitute $x=4$
- Finally put $x=4$ into the simplified expression $3x-1$: $3\cdot4-1 = 12-1 = 11$.
- That is choice (D).
- The trap $10$ in (C) is $3x-2$, what you get by forgetting the outside $+1$; the trap $12$ in (E) is $3x$, what you get if the constants $-2+1$ are wrongly treated as $0$.
💡 Once the expression is boiled down to $3x-1$, one substitution finishes it.
6.EE.A.3 Look at the first two terms: $(3x-2)(4x+1)$ and $(3x-2)\cdot4x$. Both are $(3x-2 6.EE.A.3 Inside the bracket, subtract: $(4x+1)-4x$. The $4x$ and $-4x$ cancel, leaving ju 6.EE.A.3 Now bring back the $+1$ that was waiting outside: $3x-2+1$. Combine the constant 6.EE.A.2 Finally put $x=4$ into the simplified expression $3x-1$: $3\cdot4-1 = 12-1 = 11$ Review
Reasonableness: Check the shortcut against a straight plug-in. At $x=4$: $(3\cdot4-2)=10$ and $(4\cdot4+1)=17$, so the first product is $10\cdot17=170$; the second term is $10\cdot16=160$; then $170-160+1=11$. This matches the $3x-1=11$ from the factoring shortcut, confirming (D). The answer also sits sensibly among the choices — it is not a wild number — and the near-misses $10$ and $12$ are exactly the values the two most common slips produce.
Alternative: Skip the factoring entirely and substitute $x=4$ from the start: $(10)(17)-(10)(16)+1 = 170-160+1 = 11$. It reaches (D) too, but multiplies larger numbers; the factoring route avoids the big products by cancelling $4x$ before any substitution.
CCSS standards used (min grade 6)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring the shared $(3x-2)$ out of the first two terms, cancelling $4x$ inside the bracket, and combining the constants to reach $3x-1$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $x=4$ into the simplified expression $3x-1$ to get $11$.)
⭐ When two terms share the same factor, pull it out front first — the mess often cancels and leaves a tiny expression to plug into.
⭐ When two terms share the same factor, pull it out front first — the mess often cancels and leaves a tiny expression to plug into.
More like this
Same archetype — closest grade level first.